Unformatted text preview: ty. 4.6 Lyapunov Functions
A continuous function
is a strong Lyapunov function for an equilibrium of a flow
on
if there exists an open neighborhood of such that
for
and
for all
and
.
is a weak Lyapunov function if the same holds except
and
. for all is positive definite on a neighborhood of
if
and
.A
strong Lyapunov function is positive definite in some neighborhood of an equilibrium point .
Theorem 4.7 (Lyapunov Functions) Let be an equilibrium point of a flow
. If is a
weak Lyapunov function in some neighborhood of , then is stable. If is a strong
Lyapunov function, then is asymptotically stable.
Proof. First prove that
Let
and is stable if is a weak Lyapunov function. Choose
so small that
Then
. Choose so small that
. Then if
,
for all
This means
is stable. Next prove that is asymptotically stable if is a strong Lyapunov function. Note that is
still stable as argued above. Consider a trajectory whose initial point is any
. We will
show that is the only limit point of the flow. Together with the fact that is stable, this
shows that
. 12
Chapter 4A Finally prove that is the only limit point of the flow. Consider any sequence of times
.
Then
has a convergent subsequence. For simplicity, denote the subsequence of times
too. Suppose
. Because is a decreasing function
.
However, because is not an equilibrium point,
continuity of , for for any . Fix large enough, we must also have , By . But then for , which contradicts the chain of inequalities displayed above. □
Example. Show that the origin is an asymptotically stable state for the system
,
.
For a candidate Lyapunov function, try a very simple positive definite function with minimum at
the origin. Let
. Then by the chain rule
for . is a strong Lyapunov function. The origin is an asymptotically stable equilibrium. ■
Recall Hamiltonian systems. Let
and and consider
. often represents the energy of a system. It is constant along trajectories:
.
If is an equilibrium, then Is zero at the equilibrium and constant along trajectories. If is positive definite in some
neighborhood of the equilibrium, then it is a weak Lyapunov function.
Example. The simple pendulum. The Hamiltonian is
, , leading to 13
Chapter 4A .
sketch support, rod, bob, angle is an equilibrium point. . Consider
. is positive definite in a neighborhood of the origin, so
Conclude that
is a stable equilibrium point. is a weak Lyapunov function. sketch phase portrait near origin. Nearly circular orbits surround the origin.
The linearized system is a center. ■
Example. The damped pendulum. The pendulum experiences friction proportional to velocity.
,
.
We now expect the system to lose energy. The system is no longer Hamiltonian.
an equilibrium point. Is
still a Lyapunov function? is still .
Since
when
, is still a weak Lyap nov f n tion. Lyap n...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .
 Fall '14
 MarcEvans

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